In solid sodium chloride (table salt), chloride ions have one more electron than they have protons, and sodium ions have one more proton than they have electrons. These ions are separated at about 0.28 nm. Calculate the electrostatic force between the sodium ion and the chloride ion.
What I did was the following:
F=k|q1*q2|/r^2
F=8.99e9(1.602e-19*1.602e-19)/(2.8e-10)^2
F= 2.94e-9 N
But that's not the answer. Can someone please tell me what I'm doing wrong.
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To calculate the electrostatic force between two ions, you are correct in using Coulomb's law:
F = (k * |q1 * q2|) / r^2
where F is the force, k is the Coulomb constant (8.99 × 10^9 N m^2/C^2), q1 and q2 are the charges of the two ions, and r is the distance between their centers.
However, there is an issue with the charge values you provided. The charges of the ions are not equal to the number of protons or electrons they have. In sodium chloride (NaCl), the sodium (Na) ion has a single positive charge (Na+), given that it loses one electron, while the chloride (Cl) ion has a single negative charge (Cl-), as it gains one electron.
So, to find the electrostatic force between sodium and chloride ions in solid sodium chloride, you need to adjust the charge values:
q1 = +e (where e is the elementary charge, approximately 1.602 × 10^-19 C)
q2 = -e
Now you can calculate the force:
F = (8.99 × 10^9 N m^2/C^2) * (|+e * -e|) / (0.28 × 10^-9 m)^2
F = (8.99 × 10^9) * (1.602 × 10^-19)^2 / (0.28 × 10^-9)^2
Calculating this expression will give you the correct electrostatic force between the sodium and chloride ions in solid sodium chloride.